?
The base of triangle is 15 cm and height is 12 cm the height of another triangle of double the area having base 20 cm is :
Answer & Solution
Correct Answer:
Option
C
Given base of triangle and its height is 15 cm and 12 cm respectively
Area of first triangle :
$$\eqalign{ & = \frac{1}{2} \times {\text{Base}} \times {\text{Height}} \cr & = \frac{1}{2} \times 15 \times 12 \cr & = 90{\text{ sq}}{\text{. cm}} \cr} $$
According to the question,
Let height of triangle be h cm
Area of new triangle = 180 sq. cm
Base = 20 sq. cm
$$\eqalign{ & \Rightarrow 180 = \frac{1}{2} \times 20 \times {\text{Height}} \cr & \Rightarrow {\text{h}} = \frac{{2 \times 180}}{{20}} \cr & \Rightarrow {\text{h}} = 18{\text{ cm}} \cr} $$
Area of first triangle :
$$\eqalign{ & = \frac{1}{2} \times {\text{Base}} \times {\text{Height}} \cr & = \frac{1}{2} \times 15 \times 12 \cr & = 90{\text{ sq}}{\text{. cm}} \cr} $$
According to the question,
Let height of triangle be h cm
Area of new triangle = 180 sq. cm
Base = 20 sq. cm
$$\eqalign{ & \Rightarrow 180 = \frac{1}{2} \times 20 \times {\text{Height}} \cr & \Rightarrow {\text{h}} = \frac{{2 \times 180}}{{20}} \cr & \Rightarrow {\text{h}} = 18{\text{ cm}} \cr} $$
Join the Discussion
Login to post a comment or share your explanation.
Login