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 $2=\frac{2 \times \text { Area of } \triangle\left(A_{2}\right)}{\text { Height }\left(\mathrm{H}_{2}\right)}$
$\frac{B_{1}}{B_{2}}=\frac{2 \times \frac{A_{1}}{H_{1}}}{2 \times \frac{A_{2}}{H_{2}}}$
$\frac{4 x}{3 y} \times \frac{4 y}{3x}$
$\frac{B_{1}}{B_{2}}=\frac{16}{9}$
$B_{1}: B_{2}=16: 9$
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