Class : 10 : Mensuration

Mensuration (क्षेत्रमिति)

(Plane Surface, Cylinder & Sphare, Prism & Pyramid )

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1. Plane Surface (समतलीय सतह)

Name/Figures	Area (A)	Perimeter (P)
a. समबाहु त्रिभुज(Equilateral Triangle)	$A=\frac{\sqrt{3}}{4}a^2$	$P=3a$
b. समद्विबाहु त्रिभुज(Isosceles Triangle)	$A=\frac{b}{4} \sqrt{4a^2 - b^2}$	$P=b+2a$
c. बिसमबाहु त्रिभुज(scalen Triangle)	$A=\sqrt{s(s-a)(s-b)(s-c)}$	$P=a+b+c$
d. समकाेणी त्रिभुज(Right angled Triangle)	$A=\frac{1}{2}b \times p$	P=p + b + $\sqrt{p^2+b^2}$
e. समानान्तर चतुर्भुज(Parallelogram)	$A=b \times h$	P=2(AD+AB)
f. समबाहु चतुर्भुज (Rhombus)	$A=\frac{1}{2}(d_1 \times d_2)$	$P=4a$
g. चतुर्भुज (Quadrilateral)	$A=\frac{1}{2}d(p_1 + p_2)$	P=AB+BC+CD+DA
h. समलम्ब चतुर्भुज (Trapezium)	$A=\frac{1}{2}h(p_1 + p_2)$	P=$p_1+p_2$+AB+CD
i. एराेहेड (Arrow-head)	$A=\frac{1}{2}(d_1 \times d_2)$	P=2(AB+BC) or P=2(CD+AD)
j. चङ्गा (Kite)	$A=\frac{1}{2}(d_1 \times d_2)$	P=2(AB+BC) or P=2(CD+AD)

$\large \textbf{Example: 1.}$

दिइएका चित्रहरुकाे क्षेत्रफल निकाल्नुहाेस् ।
Find the area of the given triangles.

a. b.c.

In right agled $\triangle ABC$
AB=perpendikular height (P) = 3cm
BC=base (b) =4cm
Area of the $\triangle ABC$=?
Using the formula, $Ar(\triangle ABC) = \frac{1}{2} \times b \times p \\ Ar(\triangle ABC) = \frac{1}{2} \times 4 \times 3 \\ Ar(\triangle ABC) = \frac{12}{2} \\ \therefore Ar(\triangle ABC) = 6 cm^2 $

In equilateral $\triangle ABC$,
The length of the side (a) = 5cm
Area of the $\triangle ABC$=?
Using the formula, $Ar(\triangle ABC) = \frac{\sqrt{3}}{4} a^2 \\ Ar(\triangle ABC) = \frac{\sqrt{3}}{4} \times 5^2 \\ Ar(\triangle ABC) = \frac{\sqrt{3}}{4} \times 25 \\ \therefore Ar(\triangle ABC) = \frac{25\sqrt{3}}{4}cm^2$

In scalen $\triangle ABC$,
The length of the sides: BC=a = 4.5cm, AC=b=4cm, AB=c=7cm
Area of the $\triangle ABC$=?
Using the formula, $ semi-perimeter(s)=\frac{a+b+c}{2} \\ s=\frac{4.5+4+7}{2} \\ s=\frac{15.5}{2} \\ \therefore s=7.75 cm $
and, $Ar(\triangle ABC) = \sqrt{s(s-a)(s-b)(s-c)} \\ Ar(\triangle ABC) = \sqrt{7.75(7.75-4.5)(7.75-4)(7.75-7)} \\ Ar(\triangle ABC) = \sqrt{7.75(3.25)(3.75)(0.75)}\\ Ar(\triangle ABC) = \sqrt{70.84}\\ \therefore Ar(\triangle ABC) = 8.42cm^2$

$\large \textbf{Example: 2.}$

कुनै त्रिभुजकाे एउटा भुजाकाे लम्बाइ 10cm र त्यसकाे क्षेत्रफल 60$cm^2$ छ । यदि उक्त त्रिभुज समद्विबाहु हाे भने बराबर भुजाहरुकाे नाप पत्ता लगाउनुहाेस् ।
A side of a triangle is 10cm and its area is 60$cm^2$. If the triangle is isosceles then find the measure of equal sides.

Here,

मानाै, एउटा समद्विबाहु त्रिभुजकाे बराबर भुजाहरु a र अर्काे भुजा b=10cm छ ।
उक्त त्रिभुजकाे क्षेत्रफल (A) = 60$cm^2$
अब, a काे मान पत्तालगाउँ
By the formula, $A = \frac{b}{4}\sqrt{4a^2 - b^2} \\ or, 60 = \frac{10}{4}\sqrt{4a^2 - 10^2} \\ or, 60 \times \frac{4}{10}= \sqrt{4a^2 - 100} \\ or, \frac{240}{10}= \sqrt{4a^2 - 100} \\ or, 24 = \sqrt{4a^2 - 10} \\ \text{Squaring on both sides.} \\ or, (24)^2 = (\sqrt{4a^2 - 100})^2 \\ or, 576 = 4a^2 - 100 \\ or, 576 +100 = 4a^2 \\ or, 676 = 4a^2 \\ or, \frac{676}{4} = a^2 \\ or, 169 = a^2 \\ or, \sqrt{169} = a \\ \therefore a = 13cm $
अतः बराबर भुजाहरुकाे नाप 13cm, 13cm रहेछ ।

$\large \textbf{Example: 3.}$

एउटा त्रिभुजाकार जग्गाका भुजाहरु 2:3:4 काे अनुपातमा छन् । यदि यसकाे परिमिति 900 मिटर भए, उक्त्त जग्गाकाे क्षेत्रफल कति हुन्छ? पत्ता लगाउनुहाेस् ।।
The sides of a triangular field are in the ratio 2:3:4. If its perimeter is 900 m, what is the area of that field? Find it.

Here,

मानाै, त्रिभुजाकार जग्गाकाे भुजहरु a=2x m, b=3x m र c=4x m छ ।
परिमिति (P) = 900 m
We know that, $a+b+c=P \\ or, 2x + 3x + 4x = 900 \\ or, 9x = 900 \\ or, x = \frac{900}{9} \\ \therefore x= 100 m $
भुजाहरु, $a = 2x = 2 \times 100 = 200 m\\ b= 3x = 3 \times 100 = 300 m\\ c= 4x = 4 \times 100 = 400 m $\\
Now, अर्धपरिमिति $(s) =\frac{a+b+c}{2}\\ =\frac{200+300+400}{2}\\ =\frac{900}{2}\\ =450 m$
उक्त्त जग्गाकाे क्षेत्रफल $(A) = \sqrt{s(s - a)(s - b)(s - c)}\\ = \sqrt{450(450 - 200)(450 - 300)(450 - 400)}\\ = \sqrt{450 \times 250 \times 150 \times 50}\\ = \sqrt{843750000}\\ = 29,047.38 m^2 $

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2. Cylinder and Sphere (बेलना र गाेला)

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3. Prism and Pyramid (प्रिज्म र पिरामिड)

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