Exercise 5.1 - Mensuration - Area of Plane Surfaces

[Refer to your vedanta Excel in Mathematics Book 10 for the figures.]

Creative Section A


3 a) Derive that the area of an equilateral triangle is $\frac{\sqrt{3}}{4}$ (side)².


3 b) Derive that the area of an isosceles triangle is $\frac{1}{4} \sqrt{p^2 - q^2}$.


3 c) If x, y and z are the three sides and s is the semi-perimeter of a triangle, service that its area is $\sqrt{s(s -x)(s -y)(s -x)}.

For all above questions of Question Number 3, visit this page for complete solution guide.



4. Find the area of the following figures.

a) 
Solution:

In parallelogram ABCD,
Diagonal (AC) = 15cm
AB = 14cm
BC = 13cm
In triangle ABC,
Semi-perimeter (s) = $\frac{AB + BC + AC}{2}$

$= \frac{14+13+15}{2}$
$= \frac{21} cm$

Using Heron's Formula to find the area of ∆ABC
∆ = $\sqrt{s (s - AB) (s - BC) (s - AC)}$

$= \sqrt{21 ( 21 - 14)(21-13)(21-15)}$

$= \sqrt{21 × 7 × 8 × 6}$
$= \sqrt{7056}$
$= 84 cm^2$.

And,
Area of parallelogram ABCD = 2∆
= 2×84 cm²
= 168 cm²

5 a) The adjoining figure is a trapezium ABCD. Find
  • it's area
  • the length of sides BC and AD.
Solution:
In a trapezium ABCD,
AC = 20 cm
AB = 16 cm
CD = 21 cm
<ABC = 90°
AB//CD
In right angled ∆ABC,
Using Pythagoras Theorem,
$AC^2 = AB^2 + BC^2$
$or, 20^2 = 16^2 + BC^2$
$or, BC^2 = 20^2 - 16^2$
$or, BC = \sqrt{400 - 324}$
$\therefore BC = 12 cm$
Now,
Height of trapezium (h) = BC = 12cm
One of the parallel lines ($l_1$) = 16cm
Another parallel line ($l_2$) = 21cm

Area of trapezium = $\frac{1}{2} × h(l_1 + l_2)$

$= \frac{1}{2} × 12(16+21)$
$= 6 × 37$
$= 222 cm^2$

Let a point P divide CD such that CP = AB
Now,
DP = CD - AB
= 21 - 16
= 5 cm
In right angled ∆APD,
Using Pythagoras Theorem,
AD² = DP² + AP²
or, $AD^2 = DP^2 + BC^2$ [BP = AP, height]

or, $AD^2 = 5^2 + 12^2$
$\therefore AD = 13cm$

6 a) The shape or a piece of land is a parallelogram whose adjacent sides are 12m and 9m and the corresponding diagonal is 15m. Find the area of the land.

Solution:

Let ABCD be a parallelogram.
Given,
AB = 12m
BC = 9m
AC = 15m
To find: area of parallelogram ABCD = ?

We know,
Area of parallelogram = 2 area of ∆ABC

Now,
In ∆ABC,

$s = \frac{AB + BC + AC}{2}$

$or, s = \frac{12+9+15}{2}$
$\therefore s = 18m$

Using Heron's Formula,
∆ = $\sqrt{ s(s-AB)(s - BC)(s - AC)}$

$= \sqrt{ 18(18 - 12) (18 - 9)( 18- 15)}$

$= \sqrt{18(6)(9)(3)}$
$= \sqrt{2916}$
$= 54 m^2$

Again,
Area of parallelogram ABCD = 2∆
= 2 × 54
= 108 m²

d) A garden is in the shape of a rhombus whose each side is 15 m and its one diagonals is 24 m. Find the area of the garden.

Solution:

We know,
area of a rhombus = $\dfrac{1}{2} * d1 * d2$

Let the sides of the rhombus be denoted by a and its diagonals by d1 and d2.
so, $a = 15m$ and $d1 = 24m$

midpoint of diagonal d1 = $\dfrac{d1}{2} = \dfrac{24}{2} =12m$
midpoint of diagonal d2 = $\dfrac{d2}{2}$
The midpoint of diagonals of a rhombus intersect at a right angle. So, they form a right angled triangle with a, d1/2 and d2/2.

Using Pythagorean theorem,
$\dfrac{d2}{2} = \sqrt{ a^2 - \dfrac{d1}{2}^2}$

$= \sqrt{15^2 - 12^2}$

$= \sqrt{9^2}$

$= 9m$

So, d2 = 2* 9 = 18m

And,
area of the garden in the shape of a rhombus = $\dfrac{1}{2} * 24 * 18$

$= 12*18$

$= 216 cm^2$

Hence, the required area of the garden is 216 cm^2.


7 a) If the ratio of the sides of triangle are 13:14:15 and its perimeter is 84 cm, find the area of the triangle.

Solution:
Let the sides of the triangle be 13x, 14x and 15x.

Given,
Perimeter (P) = 84cm
or, 13x + 14x + 15x = 84
or, 42x = 84
So, x = 2cm

Now,
13x = 26cm
14x = 28cm
15x = 30cm
Semi-perimeter (s) = P/2 = 84/2 = 42cm

Using Heron's Formula
Area of ∆ = $\sqrt{s(s-26)(s-28)(s-30)}$

$= \sqrt{42(42-26)(42-28)(42-30)}$

$= \sqrt{42 × 16 × 14 × 12}$

$= 336cm^2$


Creative Section - B

8 a) An umbrella is made up of 6 isosceles triangular pieces of cloths. The measurement of the base of each triangular piece is 28 cm and two equal sides are 50cm each. If the rate of cost of the cloth is Rs 0.50 per sq.cm., find the cost of the clothes required to make the umbrella.

Solution:
Here,
number of triangular faces (n) = 6

Given,
base (b) = 28cm
equal sides (a) = 50cm
Area of each ∆ = $\frac{b}{4} \sqrt{4a^2 - b^2}$

$= \frac{28}{4} \sqrt{4 × 50^2 - 28^2}$

$= 7 × 96$
$= 672 cm^2$
And,
Total required area = n×∆
= 6 × 672cm²
= 4032 cm²
Also,
rate = Rs 0.50/cm²
Required total cost = 0.50 × 4032
= Rs 2016


About vedanta EXCEL in MATHEMATICS Book 10

Author: Hukum Pd. Dahal
Editor: Tara Bahadur Magar

Vanasthali, Kathmandu, Nepal
+977-10-4382404, 01-4362082
vedantapublication@gmail.com

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Class 10 - Mensuration - Area of Plane Surfaces Solved Exercises | vedanta Excel in Mathematics is a collection of the solutions related to exercises of area of plane surfaces from Mensuration Chapter for Nepal's Secondary Education Examination (SEE) appearing students.

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