SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14D

 Exercise 14 D 

Question 9 

A rectangular field is 24 m long and 15 m wide. How many triangular flower beds each of base 3m and altitude 4m can be laid in this field? 

Sol: No of triangular flower =  $\frac{Area of rectangular field}{Area of triangular flower}$

$=\frac{l \times B}{\frac{1}{2} \times \text { Base } \times \text { Height }}$

$=\frac{24 \times 15}{\frac{1}{2} \times 3 \times 4}=60$

Question 10

A right angled triangle has the largest side as 13 cm and one of the sides containing the right angle as 12cm. It's area in $\mathrm{Cm}^{2}$ is :

(IMAGE TO BE ADDED)

Sol:

 $\begin{aligned} H^{2} &=B^{2}+p^{2} \\ \Rightarrow P^{2} &=H^{2}-B^{2} \\ &=13^{2}-12^{2}=169-144=25 \end{aligned}$

$\Rightarrow p^{2}=25^{\circ}$

$\because \quad p=5 \mathrm{~cm}$

Area of Triangle = $\frac{1}{2} \times$ Base $\times$ height $=\frac{1}{2} \times 12 \times 5$

= 30 $\mathrm{Cm}^{2}$

Question 11

The area of a right angled triangle is 40 times its base. what is its height ? 

Sol: Area = $40 \times B$ 

Area of Triangle = $\frac{1}{2} \times$ Base $\times$ Height

Height $=\frac{2 \times \text { Area }}{\text { Base }}=\frac{2 \times 40 \times B}{B}$

= 80 cm 

Question 12

In two triangles, the ratio of their areas is 4:3 and that of their height is 3:4 .find the ratio of their bases.

Sol: Ratio of areas of triangles $A_{1}: A_{2}$ is $4 x: 3 u$

Ratio of their height $H_{1}: H_{2}$ is $3 y: 4 y$

now area of triangle = $\frac{1}{2} \times$ Base $\times$ Height

Base $B_{1}=\frac{2 \times \text { Area of } \Delta\left(A_{1}\right)}{\text { Height }\left(H_{1}\right)}$

Basc B $2=\frac{2 \times \text { Area of } \triangle\left(A_{2}\right)}{\text { Height }\left(\mathrm{H}_{2}\right)}$

$\frac{B_{1}}{B_{2}}=\frac{2 \times \frac{A_{1}}{H_{1}}}{2 \times \frac{A_{2}}{H_{2}}}$

$\frac{4 x}{3 y} \times \frac{4 y}{3x}$

$\frac{B_{1}}{B_{2}}=\frac{16}{9}$

$B_{1}: B_{2}=16: 9$