Question
A triangle and a parallelogram have the same base and the same area. If the sides of triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Solution:
For triangle
Perimeter of triangle $=(26+28+30) \mathrm{cm}=84 \mathrm{~cm}$
$2 s=84 \mathrm{~cm}$
$s=42 \mathrm{~cm}$
By Heron's formula,
Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
Area of triangle $=[\sqrt{42(42-26)(42-28)(42-30)}] \mathrm{cm}^{2}$
$=[\sqrt{42(16)(14)(12)}] \mathrm{cm}^{2}=336 \mathrm{~cm}^{2}$
Let the height of the parallelogram be h.
Area of parallelogram = Area of triangle
$h \times 28 \mathrm{~cm}=336 \mathrm{~cm}^{2}$
$h=12 \mathrm{~cm}$
Therefore, the height of the parallelogram is $12 \mathrm{~cm}$.
Video Solution for heron s formula (Page: 206 , Q.No.: 4)
For triangle
Perimeter of triangle $=(26+28+30) \mathrm{cm}=84 \mathrm{~cm}$
$2 s=84 \mathrm{~cm}$
$s=42 \mathrm{~cm}$
By Heron's formula,
Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
Area of triangle $=[\sqrt{42(42-26)(42-28)(42-30)}] \mathrm{cm}^{2}$
$=[\sqrt{42(16)(14)(12)}] \mathrm{cm}^{2}=336 \mathrm{~cm}^{2}$
Let the height of the parallelogram be h.
Area of parallelogram = Area of triangle
$h \times 28 \mathrm{~cm}=336 \mathrm{~cm}^{2}$
$h=12 \mathrm{~cm}$
Therefore, the height of the parallelogram is $12 \mathrm{~cm}$.
Video Solution for heron s formula (Page: 206 , Q.No.: 4)