एउटा जग्गा 120 मिटर लम्बाइ र 60 मिटर चौडाइ भएको छ । जग्गाको बिचमा एउटा 12 मिटर भुजा भएको समबाहु त्रिभुज आकारको बगैंचा छ ।
(In a land of length 120 m and breadth 60 m, there is a garden in the shape of an equilateral triangle of side 12 m.)
उक्त आयताकार जग्गाको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the rectangular land.)[1U]
उक्त बगैंचाको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the garden.)[1U]
बगैंचाबाहेक उक्त जग्गाको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the land excluding the garden.)[1A]
उक्त बगैंचामा प्रतिमिटरको रु. 500 को दरले बार लगाउन कति खर्च लाग्छ पत्ता लगाउनुहोस् ।
(Find the cost of fencing the garden at the rate of Rs. 500 per meter.)[2A]
The area of rectangle is \(A_1=\) length × breadth
or \(A_1= 120 \times 60 = 7200 \text{m}^2\)
The garden is an equilateral triangle with side \( 12 \text{m} \). Therefore, the area of equilateral triangle is \(A_2= \frac{\sqrt{3}}{4} \times (\text{side})^2 \)
or \(A_2=\frac{\sqrt{3}}{4} \times 12^2 = \frac{\sqrt{3}}{4} \times 144 = 36\sqrt{3} =62.35 \text{m}^2\)
The area of the land excluding the garden is \(A=A_2-A_1\)
or \(A= 7200 - 62.35 =7137.65 \text{m}^2 \)
The perimeter of equilateral triangle is \(P= 3 \times \text{side} = 3 \times 12 = 36 \text{m} \)
Henvce, the cost of fencing the garden at the rate of Rs. 500 per meter is Cost of fencing = P × Rate
or Cost of fencing = \(36 \times 500 = Rs 18000\)
दिइएको चित्रमा ABCD एउटा वर्ग हो र त्यसभित्र एउटा वृत्त खिँचिएको छ ।
(In the given figure, ABCD is a square and a circle is drawn inside it.)
वर्गको एउटा भुजा AB = 14 cm भए, यसको क्षेत्रफल पत्ता लगाउनुहोस् ।
(If one side of the square AB = 14 cm, find its area.)[1U]
वृत्तको अर्धव्यास कति रहेछ ?
(How much is the radius of the circle?)[1K]
छाया परिएको भागको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the shaded portion.)[2A]
उक्त वृत्तको परिधिको लम्बाइ पत्ता लगाउनुहोस् ।
(Find the length of the circumference of the circle.)[1HA]
The area of the square is \(A_1 = l^2\)
or \(A_2 = 14^2 = 196 \text{cm}^2\)
Since the circle is inscribed in the square, its diameter equals the side of the square. Diameter \((d)= 14 \text{cm}\)
So, Radius \((r)= 7 \text{cm}\)
The shaded portion is the area of the square minus the area of the circle. Area of circle \(A_2= \pi r^2 = \pi \times 7^2 = 49\pi \text{cm}^2\)
SO, area of shaded area is Shaded area \(A= A_1 - A_2 = 196 - 49\pi=196 - 153.94 = 42.06 \text{cm}^2\)
The circumference of the circle is \(C = 2\pi r = 2 \pi \times 7 = 14\pi = 14\pi \approx 43.98 \text{cm}\)
एउटा वृत्ताकार पोखरीको क्षेत्रफल 616 m² छ ।
(The area of a circular pond is 616 m².)
उक्त पोखरीको अर्धव्यास पत्ता लगाउनुहोस् ।
(Find the radius of the pond.)[2U]
उक्त पोखरीको परिधिको लम्बाइ पत्ता लगाउनुहोस् ।
(Find the length of the circumference of the pond.)[1A]
उक्त पोखरीलाई पाँच फन्को तारले घेर्न कति मिटर तार चाहिन्छ ?
(How much wire is required to surround the pond 5 times by the wire?)[1A]
प्रतिमिटरको रु. 300 को दरले उक्त पोखरीमा बार लगाउन जम्मा कति खर्च लाग्छ ?
(How much does it cost for fencing the pond at the rate of Rs. 300 per meter?)[1HA]
The area of the circular pond is given as \(616 \text{m}^2\).Therefore \(A = \pi r^2 = 616\)
or \(\dfrac{22}{7} r^2 = 616\)
or \(r^2 = 616 \times \dfrac{7}{22} = 196\)
or \(r = \sqrt{196} = 14 \text{m}\)
The circumference of the pond is \(C = 2\pi r \)
or \(C=2 \times \dfrac{22}{7} \times 14= 88 \text{m}\)
To surround the pond 5 times, total wire required is Length of wire \(= 5 \times C = 5 \times 88= 440 \text{m}\)
Cost of fencing at Rs. 300 per meter is Cost \(= C \times 300 = 88 \times 300\)
एउटा वृत्ताकार जग्गाको क्षेत्रफल 2464 वर्गमिटर छ ।
(The area of a circular land is 2464 square meters.)[2U]
उक्त जग्गाको अर्धव्यास पत्ता लगाउनुहोस् ।
(Find the radius of the land.)[2A]
उक्त जग्गाको परिधिको लम्बाइ पत्ता लगाउनुहोस् ।
(Find the length of the circumference of the land.)[1U]
उक्त जग्गामा प्रतिमिटरको रु. 1200 का दरले बार लगाउन कति खर्च लाग्छ ? पत्ता लगाउनुहोस् ।
(Find the cost of fencing the land at the rate of Rs. 1200 per meter.)[1A]
उक्त जग्गामा लम्बाइ 12 m भएको एउटा वर्गाकार पोखरी रहेछ भने उक्त पोखरीको क्षेत्रफल पत्ता लगाउनुहोस् ।
(If there is a square pond of length 12 m in the land, find the area of the pond.)[1U]
The area of the circular land is given as \(2464 \text{m}^2\). Therefore \(A = \pi r^2 = 2464\)
or \(\dfrac{22}{7} r^2 = 2464\)
or \(r^2 = 784\)
or \(r = \sqrt{784} = 28 \text{m}\)
The circumference of the land is \(C = 2\pi r \)
or \(C = 2 \times \dfrac{22}{7} \times 28 = 176 \text{m}\)
Cost of fencing at Rs. 1200 per meter is Cost \(= C \times 1200 = 176 \times 1200\)
or Cost \(= \text{Rs } 211200\)
The pond is square with side \(12 \text{m}\). Therefore, its area is Area \(= (l^2 = 12^2=144 \text{m}^2\)
सँगै दिइएको चित्रमा ABCD एउटा आयत हो र वृत्तको अर्धव्यास OC हो ।
(In the adjoining figure, ABCD is a rectangle and OC is the radius of the circle.)
यदि AB = 8 cm र BC = 6 cm भए, आयतको परिमिति कति हुन्छ ?
(If AB = 8 cm and BC = 6 cm, what is the perimeter of the rectangle?)[1U]
यदि OC = 5 cm भए, वृत्तको व्यासको लम्बाइ कति हुन्छ ?
(If OC = 5 cm, what is the length of the diameter of the circle?)[1K]
चित्रमा छाया परिएको भागको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the shaded portion in the figure.)[2A]
वृत्तको परिधि कति से.मि. रहेछ, पत्ता लगाउनुहोस् ।
(Find the circumference of the circle in cm.)[1A]
Given AB = 8 cm and BC = 6 cm.
So, the perimeter of the rectangle is P \(= 2(l + b)\)
or P \(= 2(8 + 6) = 2 \times 14 = 28 \text{cm}\)
Given OC = 5 cm, which is the radius of the circle.
Therefore, diameter of the circle is d \(= 2 \times \text{radius} = 2 \times 5\)
or d \(= 10 \text{cm}\)
The shaded portion is the area of the circle minus the area of the rectangle.
Here, the area of circle is
\(A_1 = \pi r^2 = \dfrac{22}{7} \times 5^2 = \dfrac{22}{7} \times 25 = \dfrac{550}{7} = 78.57 \text{cm}^2\)
Next, the area of rectangle is \(A_2 = l \times b = 8 \times 6 = 48 \text{cm}^2\)
SO, area of shaded portion is Shaded area \((A)= A_1 - A_2 = 78.57 - 48 = 30.57 \text{cm}^2\)
The circumference of the circle is \(C = 2\pi r = 2 \times \dfrac{22}{7} \times 5= 31.43 \text{cm}\)
सँगै दिइएको चित्र एउटा समलम्ब चतुर्भुज ABCD आकारको एक टुक्रा जग्गा छ । जसमा AB//DC, BC⊥AB छ ।
(In the given figure, ABCD is a piece of land in the shape of a trapezium.)
उक्त जग्गाको क्षेत्रफल वर्गमिटरमा पत्ता लगाउनुहोस् ।
(Find the area of the land in square meter.)[1U]
उक्त जग्गामा प्रतिमिटरको रु. 200 का दरले बार लगाउन कति खर्च लाग्छ ? पत्ता लगाउनुहोस् ।
(Find the cost of fencing the land at Rs. 200 per meter.)[2A]
त्रिभुजाकार भाग ABE को क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the triangular part ABE.)[1U]
छाया परिएको भागको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the shaded portion.)[1A]
ABCD is a trapezium with AB ∥ DC and BC ⊥ AB.
So, the area of the land is Area of trapezium \(= \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
or Area of trapezium \(= \dfrac{1}{2} \times (10 + 15) \times 6 = \dfrac{1}{2} \times 25 \times 6 = 75 \text{m}^2\)
To find fencing cost, first find perimeter of trapezium ABCD. So, the perimeter is Perimeter \((P)= 10 + 8 + 15 + 12 = 45 \text{m}\)
Cost at Rs. 200 per meter, so the cost is Cost \(= 45 \times 200Rs 9000\)
Triangle ABE has base AB = 10 m and height = 6 m , so its area ia Area of \(\triangle ABE = \dfrac{1}{2} \times \text{base} \times \text{height}\)
or Area of \(\triangle ABE = \dfrac{1}{2} \times 10 \times 6 = 30 \text{m}^2\)
The area of shaded portion is A=Total area of trapezium – Area of unshaded triangle ABE
orA\(= 75 - 30 = 45 \text{m}^2\)
सँगैको चित्रमा ABCD एउटा वर्गाकार पार्क हो । उक्त पार्कभित्र एउटा वृत्ताकार पोखरी छ ।
(In the adjoining figure, ABCD is a square park. Inside that park, there is a circular pond.)
उक्त पार्कको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the park.)[1U]
उक्त वृत्ताकार पोखरीको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the circular pond.)[1U]
पोखरीबाहेक पार्कको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the park excluding the pond.)[1A]
उक्त पार्कमा रु. 650 प्रतिमिटरको दरले बार लगाउन जम्मा कति खर्च लाग्छ ? पत्ता लगाउनुहोस् ।
(How much does it cost to fence the park at the rate of Rs. 650 per meter? Find it.)[2A]
The park is a square with side AB = 25 m. So, are aof park is Area of park \(A_1 = l^2= 25^2 = 625 \text{m}^2\)
The circular pond has a diameter of 14 m , so
So, radius is radius \(r = \dfrac{14}{2} = 7 \text{m}\).
Now, area of circular pond is Area of circular pond \(A_2 = \pi r^2 = \dfrac{22}{7} \times 7^2\)
or \(A_2 = \dfrac{22}{7} \times 49 = 154 \text{m}^2\)
Area of park excluding the pond is \(A = A_1 - A_2 = 625 - 154=471 \text{m}^2\)
The perimeter of the paark is P \(= 4 \times l = 4 \times 25 = 100 \text{m}\)
So,to fence the park, at Rs. 650 per meter, the cost is Cost \(= 100 \times 650=Rs65000\)
20 m लम्बाइ र 15 m चौडाइ भएको एउटा टुक्रा आयताकार जग्गाभित्र 7 m व्यास भएको एउटा वृत्ताकार पोखरी छ ।
(A circular pond of diameter 7 m is inside a rectangular piece of land of length 20 m and breadth 15 m.)
उक्त जग्गाको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the land.)[1U]
उक्त पोखरीको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the pond.)[1U]
पोखरीबाहेक जग्गाको क्षेत्रफल कति रहेछ, पत्ता लगाउनुहोस् ।
(Find the area of the land excluding the pond.)[1A]
उक्त जग्गामा प्रतिमिटरको रु. 175 का दरले बार लगाउन लाग्ने खर्च कति लाग्छ ? पत्ता लगाउनुहोस् ।
(Find the cost of fencing the land at the rate of Rs. 175 per meter.)[2A]
The land is a rectangle with length = 20 m and breadth = 15 m. So, area of land is Area of land \(A_1 = l \times b = 20 \times 15 = 300 \text{m}^2\)
The circular pond has a diameter of 7 m, so
its radius is radius \(r = \dfrac{7}{2} = 3.5 \text{m}\).
Now, area of circular pond is Area of pond \(A_2 = \pi r^2 = \dfrac{22}{7} \times (3.5)^2\)
or \(A_2 = \dfrac{22}{7} \times 12.25 = 38.5 \text{m}^2\)
Area of land excluding the pond is \(A = A_1 - A_2 = 300 - 38.5 = 261.5 \text{m}^2\)
The perimeter of the land is P \(= 2(l + b) = 2(20 + 15) = 2 \times 35 = 70 \text{m}\)
So, to fence the land at Rs. 175 per meter, the cost is Cost \(= 70 \times 175 = \text{Rs } 12250\)
चित्रमा एउटा वर्ग ABCD भित्र एउटा वृत्त देखिएको छ ।
(In the figure, a circle is enclosed in a square ABCD.)
वृत्त र वर्गको क्षेत्रफल निकाल्ने सूत्रहरू लेख्नुहोस् ।
(Write the formula for finding the areas of a circle and a square.)[1K]
वृत्तको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the circle.)[1U]
छाया परिएको भागको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the shaded portion.)[2A]
वर्गको परिमिति कति हुन्छ ? गणना गर्नुहोस् ।
(How much is the perimeter of the square? Calculate it.)[1HA]
Formula for area of a square and a circle are: Area of square \(= l^2\) Area of circle \(= \pi r^2\)
From the figure, side of square AB = 7 cm.
Since the circle is inscribed, its diameter = side of square = 7 cm.
So, radius of the circle is \(r = \dfrac{7}{2} = 3.5 \text{cm}\)
Now, area of circle is Area of circle \(= \pi r^2 = \dfrac{22}{7} \times (3.5)^2\)
or Area of circle \(= \dfrac{22}{7} \times 12.25 = 38.5 \text{cm}^2\)
Here Area of square \(A_1 = l^2 = 7^2 = 49 \text{cm}^2\)
Again Area of circle \(A_2 = 38.5 \text{cm}^2\)
So, area of shaded portion is \(A = A_1 - A_2 = 49 - 38.5 = 10.5 \text{cm}^2\)
The perimeter of the square is P \(= 4 \times l = 4 \times 7 = 28 \text{cm}\)
चित्रमा देखिएको समलम्ब चतुर्भुज आकारको जग्गा राधाको हो । यो जग्गाभित्र एउटा वर्गाकार पोखरी पनि छ ।
(The trapezium-shaped land shown in the figure belongs to Radha. Inside the land there is a pond in the shape of a square.)
समलम्ब चतुर्भुजको क्षेत्रफल पत्ता लगाउने सूत्र लेख्नुहोस् ।
(Write the formula to find the area of a trapezium.)[1K]
पोखरीको सतहको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the surface of the pond.)[1U]
पोखरीबाहेक जग्गाको क्षेत्रफल पत्ता लगाउनुहोस् ।
(Find the area of the land excluding the pond.)[2A]
उक्त जग्गामा 2 फन्को बार लगाउन कति मिटर तार चाहिन्छ ? गणना गर्नुहोस् । How much wire is required to fence the land two rounds? Calculate it. [1HA]
Formula to find the area of a trapezium is: Area \(= \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
The pond is square with side = 4 m.
So, area of pond is Area of pond \((A_1)= l^2 = 4^2 = 16 \text{m}^2\)
From the figure, we know that
Area of trapezium land is Area of trapezium \((A_2) = \dfrac{1}{2} \times (30 + 20) \times 12 = \dfrac{1}{2} \times 50 \times 12 = 300 \text{m}^2\)
So, area of land excluding pond is \(A = A_2 - A_1 = 300 - 16 = 284 \text{m}^2\)
The perimeter of the land is Perimeter \(= 30 + 15 + 20 + 14 = 79 \text{m}\)
Now, to find wire for 2 rounds, the required wire is Total wire required \(= 2 \times 79 = 158 \text{m}\)
चित्रमा देखाइएको एउटा चतुर्भुज आकारको जग्गा सागरको हो । उक्त जग्गामा एउटा \(1.4\) मी. व्यास भएको कुवा निर्माण गरिएको छ । A piece of land in the shape of a quadrilateral shown in the figure belongs to Sagar. There is a well of diameter \(1.4\) m inside the land.
चतुर्भुजको क्षेत्रफल निकाल्ने सूत्र लेख्नुहोस् । Write the formula to find the area of a quadrilateral. [1K]
उक्त जग्गाको क्षेत्रफल पत्ता लगाउनुहोस् । Find the area of the land. [1U]
कुवाबाहेक जग्गाको क्षेत्रफल कति हुन्छ ? पत्ता लगाउनुहोस् । Find the area of the land excluding the well. [2A]
उक्त इनारको घेराको परिधि कति हुन्छ ? पत्ता लगाउनुहोस् । How much is the circumference of the lid of the well? [1HA]
Formula to find the area of a quadrilateral (when a diagonal and perpendicular heights are given) is: Area \(= \dfrac{1}{2} \times \text{diagonal} \times (h_1 + h_2)\)
From the figure, the area of quadrilateral is \(A_1 = \dfrac{1}{2} \times 20 \times (6 + 8) = \dfrac{1}{2} \times 20 \times 14 = 140 \text{m}^2\)
In the figure, given that diameter of well = 1.4 m, so radius is \(r = \dfrac{1.4}{2} = 0.7 \text{m}\)
Now, area of well is \(A_2 = \pi r^2 = \dfrac{22}{7} \times (0.7)^2 = \dfrac{22}{7} \times 0.49 = 1.54 \text{m}^2\)
Hence, area of land excluding well is \(A = A_1 - A_2 = 140 - 1.54 = 138.46 \text{m}^2\)
Circumference of the well is \(C = \pi d = \dfrac{22}{7} \times 1.4\)
or \(C = 4.4 \text{m}\)
एउटा वृत्ताकार पोखरीको सतहको क्षेत्रफल \(1386\) m\(^2\) छ । The surface area of a circular pond is \(1386\) m\(^2\).
वृत्तको क्षेत्रफल पत्ता लगाउने सूत्र लेख्नुहोस् । Write the formula to find the area of a circle. [1K]
उक्त पोखरीको व्यास पत्ता लगाउनुहोस् । Find the diameter of the pond. [2U]
उक्त पोखरीमा बार लगाउन कति मिटर तार चाहिन्छ ? पत्ता लगाउनुहोस् । How many meters of wire gauze is required to fence the pond? Find it. [1HA]
प्रतिमिटर रु. \(250\) को दरले तार जाली लगाउन कति खर्च लाग्छ ? पत्ता लगाउनुहोस् । How much does it cost to fence the pond at the rate of Rs. \(250\) per meter of wire gauze? Find it. [1HA]
Formula to find the area of a circle is: Area \(= \pi r^2\)
Given area of pond \(= 1386 \text{m}^2\).
So, \(\pi r^2 = 1386\)
or \(\dfrac{22}{7} r^2 = 1386\)
or \(r^2 = 1386 \times \dfrac{7}{22} = 441\)
or \(r = \sqrt{441} = 21 \text{m}\)
Therefore, diameter is \(d = 2r = 2 \times 21 = 42 \text{m}\)
Length of wire required to fence the pond is the circumference of the pond, so \(C = \pi d = \dfrac{22}{7} \times 42\)
or \(C = 132 \text{m}\)
Cost of fencing at Rs. 250 per meter is Cost \(= 132 \times 250\)
or Cost \(= \text{Rs } 33{,}000\)
एक धावक \(2200\) मी. लम्बाइ भएको वृताकार भागमा \(5\) फन्को मारी \(2200\) मी. दौड पूरा गर्छ । A runner running in a circular track completes \(5\) rounds and cover a distance of \(2200\) m.
वृत्तको परिमिति पत्ता लगाउने सूत्र लेख्नुहोस् । Write the formula to find the circumference of a circle. [1K]
उक्त धावमार्गको एक फन्को (round) कति लामो होला ? How much is the length of one round of the track. [1U]
उक्त धावमार्गको घेराको वृत्तको अर्धव्यासको लम्बाइ पत्ता लगाउनुहोस् । Find the length of the radius of the circle surrounded by the track. [2A]
उक्त वृत्तको क्षेत्रफल पत्ता लगाउनुहोस् । Find the area of the circle. [1U]
Formula to find the circumference of a circle is: Circumference \(= 2\pi r\)
Total distance covered in 5 rounds = 2200 m.
So, length of one round is circumference of the circle, which is \(C= \dfrac{2200}{5} = 440 \text{m}\)
Here, circumference of circle is 440 m.
So, \(2\pi r = 440\)
or \(2 \times \dfrac{22}{7} \times r = 440\)
or \(\dfrac{44}{7} r = 440\)
or \(r = 440 \times \dfrac{7}{44} = 70 \text{m}\)
Area of the circle is Area \(= \pi r^2 = \dfrac{22}{7} \times 70^2\)
or \(= \dfrac{22}{7} \times 4900 = 15400 \text{m}^2\)
चित्रमा एउटा अर्ध वृत्ताकार बगैंचा देखाइएको छ । जसको परिमिति \(72\) मी छ । The figure in a semi-circular garden is shown, whose perimeter is \(72\) m.
अर्धवृत्तको परिमिति निकाल्ने सूत्र लेख्नुहोस् । Write the formula to find perimeter of a semi-circle. [1K]
उक्त बगैंचाको अर्धव्यास पत्ता लगाउनुहोस् । Find the radius of the garden. [1U]
उक्त बगैंचाको क्षेत्रफल पत्ता लगाउनुहोस् । Find the area of the garden. [2A]
प्रति वर्गमिटर रु. \(150\) का दरले उक्त बगैंचा सम्याउने कति खर्च लाग्छ ? पत्ता लगाउनुहोस् । Find the cost of levelling the garden at the rate of Rs. \(150\) per square meter. [1HA]
Formula to find the perimeter of a semi-circle is Perimeter \(= \pi r + 2r = r(\pi + 2)\)
Given perimeter of semi-circular garden = 72 m.
So, \(r(\pi + 2) = 72\)
or\(r\left(\dfrac{22}{7} + 2\right) = 72\)
or\(r\left(\dfrac{22 + 14}{7}\right) = 72\)
or\(r \times \dfrac{36}{7} = 72\)
or \(r = 72 \times \dfrac{7}{36} = 14 \text{m}\)
Area of semi-circular garden is Area \(= \dfrac{1}{2} \pi r^2 = \dfrac{1}{2} \times \dfrac{22}{7} \times 14^2\)
or \(= \dfrac{1}{2} \times \dfrac{22}{7} \times 196 = \dfrac{1}{2} \times 616 = 308 \text{m}^2\)
Cost of levelling at Rs. 150 per m² is Cost \(= 308 \times 150\)
or Cost \(= \text{Rs } 46200\)
चित्रमा एउटा आयताकार खेत देखाइएको छ । त्यस खेतको एउटा कुनामा एउटा वर्गाकार टहरो बनाइएको छ । In the figure a rectangular field is shown. In one corner of the field a cottage in the shape of a square is made.
आयतको क्षेत्रफल पत्ता लगाउने सूत्र लेख्नुहोस् । Write the formula to find the area of a rectangle. [1K]
उक्त टहरोको क्षेत्रफल पत्ता लगाउनुहोस् । Find the area of the cottage. [1U]
टहरोबाहेक उक्त खेतको क्षेत्रफल पत्ता लगाउनुहोस् । Find the area of the field excluding the cottage. [2A]
प्रतिमिटरको रु. \(50\) का दरले उक्त खेतमा बार लगाउन कति खर्च लाग्छ ? पत्ता लगाउनुहोस् । How much does it cost to fence the field at the rate of Rs. \(50\) per meter? Find it. [1HA]
Formula to find the area of a rectangle is Area \(= l \times b\)
The cottage is square with side = 5 m.
So, area of cottage is Area \(A_1= l^2 = 5^2 = 25 \text{m}^2\)
Area of rectangular field is \(A_2 = 50 \times 30 = 1500 \text{m}^2\)
So, area of field excluding cottage is \(A = A_2 - A_1 = 1500 - 25 = 1475 \text{m}^2\)
Perimeter of the rectangular field is P \(= 2(l + b) = 2(50 + 30) = 2 \times 80 = 160 \text{m}\)
Cost of fencing at Rs. 50 per meter: Cost \(= 160 \times 50 = \text{Rs } 8000\)
चित्रमा एउटा समानान्तर चतुर्भुज देखाइएको छ । उक्त समानान्तर चतुर्भुजको एउटा कुनामा समबाहु चतुर्भुजको आकारको पोखरी छ । Figure I and in the shape of a parallelogram is shown. In one corner of the land there is a pond in the shape of a rhombus.
समबाहु चतुर्भुज (Rhombus) को विकर्ण दिइएको अवस्थामा क्षेत्रफल पत्ता लगाउने सूत्र लेख्नुहोस् । Write the formula to find the area of a rhombus when two diagonals are given. [1K]
उक्त जग्गाको क्षेत्रफल पत्ता लगाउनुहोस् । Find the area of the land. [1U]
पोखरीबाहेक उक्त खेतको क्षेत्रफल पत्ता लगाउनुहोस् । Find the area of the land excluding the pond. [2A]
प्रतिमिटरको रु. \(100\) को दरले उक्त जग्गामा बार लगाउन कति खर्च लाग्ला ? गणना गर्नुहोस् । Find the cost of fencing the land at the rate of Rs. \(100\) per meter. [1A]
Formula to find the area of a rhombus when two diagonals are given is: Area \(= \dfrac{1}{2} \times d_1 \times d_2\)
The land is a parallelogram with base = 10 m and height = 6 m (shown by dashed perpendicular).
So, area of land is Area \(A_1= \text{base} \times \text{height} = 10 \times 6 = 60 \text{m}^2\)
The pond is a rhombus with diagonals \(d_1 = 4 \text{m}\) and \(d_2 = 3 \text{m}\).
So, area of pond is \(A_2 = \dfrac{1}{2} \times 4 \times 3 = 6 \text{m}^2\)
Now, area of land excluding pond is \(A = 60 - 6 = 54 \text{m}^2\)
Perimeter of the parallelogram is P \(= 2(10+8) = 36 \text{m}\)
Cost of fencing at Rs. 100 per meter: Cost \(= 36 \times 100= \text{Rs } 3600\)
एउटा आयताकार कोठाको लम्बाइ \(20\) फिट छ र परिमिति \(70\) फिट छ । The length of a rectangular room is \(20\) ft and the perimeter is \(70\) ft.
आयतको परिमिति पत्ता लगाउने सूत्र लेख्नुहोस् । Write the formula to find perimeter of a rectangle. [1K]
उक्त कोठाको चौडाइ पत्ता लगाउनुहोस् । Find the breadth of the room. [1U]
उक्त कोठामा प्रति वर्गफिटको रु. \(200\) का दरले कार्पेट ओछ्याउन कति खर्च लाग्छ ? गणना गर्नुहोस् । How much does it cost to carpet the room at the rate of Rs. \(200\) per square feet? Calculate. [2A]
उक्त कोठाको लम्बाइ र चौडाइको अनुपात पत्ता लगाउनुहोस् । Find the ratio of the length and the breadth of the room. [1HA]
Formula to find the perimeter of a rectangle is: Perimeter \(= 2(l + b)\)
Given: length \(l = 20\) ft, perimeter = 70 ft.
So, \(2(l + b) = 70\)
or \(2(20 + b) = 70\)
or \(20 + b = 35\)
or \(b = 35 - 20 = 15 \text{ft}\)
Area of the room is \(A= l \times b = 20 \times 15 = 300 \text{ft}^2\)
Cost of carpeting at Rs. 200 per sq. ft Cost \(= 300 \times 200\)
or Cost \(= \text{Rs } 60000\)
the ratio of the length and the breadth of the room is \(20 : 15 \), which is \(4 : 3\)
एउटा \(120\) मी लम्बाइ भएको बर्गाकार बगैंचाभित्र एउटा \(18\) मी लामो र \(9\) मी चौडा भलिबल कोर्ट बनाइएको छ । A volleyball court of length \(18\) m and breadth \(9\) m is made inside a square garden of length \(120\) m.
उक्त बगैंचाको क्षेत्रफल पत्ता लगाउनुहोस् । Find the area of the garden. [1U]
उक्त भलिबल कोर्टको क्षेत्रफल पत्ता लगाउनुहोस् । Find the area of the volleyball court. [1U]
भलिबल कोर्टबाहेक उक्त वर्गको क्षेत्रफल पत्ता लगाउनुहोस् । Find the area of the garden excluding the volleyball court. [1A]
प्रतिमिटरको रु. \(250\) का दरले उक्त वर्गलाई बार लगाउन कति खर्च लाग्छ, पत्ता लगाउनुहोस् । Find the cost of fencing the garden at the rate of Rs. \(250\) per meter. [2A]
The garden is square with side = 120 m.
So, area of garden is Area \(A_1 = l^2 = 120^2 = 14{,}400 \text{m}^2\)
The volleyball court is rectangular with length = 18 m and breadth = 9 m.
So, area of court is Area \(A_2 = l \times b = 18 \times 9 = 162 \text{m}^2\)
Area of garden excluding the court is \(A = A_1 - A_2 = 14{,}400 - 162 = 14{,}238 \text{m}^2\)
Perimeter of the square garden is P \(= 4 \times l = 4 \times 120 = 480 \text{m}\)
Cost of fencing at Rs. 250 per meter: Cost \(= 480 \times 250 = \text{Rs } 120000\)
एउटा बर्गाकार खेतको परिमिति \(240\) फिट छ । The perimeter of a land in the shape of a square is \(240\) feet.
वर्गको परिमिति पत्ता लगाउने सूत्र लेख्नुहोस् । Write the formula to find the perimeter of a square. [1K]
उक्त जग्गाको लम्बाइ पत्ता लगाउनुहोस् । Find the length of the land. [1U]
उक्त जग्गा रु. \(30\) प्रति बर्गफिदका दरले सम्याउने लाग्ने खर्च पत्ता लगाउनुहोस् । Find the cost of levelling the land at the rate of Rs. \(30\) per square feet. [2A]
प्रति फिटको रु. \(50\) का दरले उक्त जग्गामा बार लगाउने कति खर्च लाग्छ ? पत्ता लगाउनुहोस् । How much does it cost to fence the land at the rate of Rs. \(50\) per feet? Find it. [1A]
Formula to find the perimeter of a square is Perimeter \(= 4 \times l\)
Given perimeter = 240 feet.
So, \(4l = 240\)
or \(l = \dfrac{240}{4} = 60 \text{feet}\)
Area of the square land \(= l^2 = 60^2 = 3600 \text{ft}^2\)
Cost of levelling at Rs. 30 per sq. ft: Cost \(= 3600 \times 30\)
or Cost \(= \text{Rs } 108000\)
Perimeter of the land = 240 feet
Cost of fencing at Rs. 50 per foot: Cost \(= 240 \times 50 = \text{Rs } 12000\)
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