A field in the form of a parallelogram has an area of 432 m - - 2 - Find the length of its diagonal on which the perpendiculars drawn from two vertices on either side of it are 12m long-

Question

Toppr Admin · Accepted Answer

Area of the triangle -12-xD7-b-xD7-h-12-xD7-x-xD7-h1-12-xD7-x-xD7-h2-432-12-xD7-x-xD7-12-12-xD7-x-xD7-12-43212x-432x-36Hence- the length of diagonal is 36 m-

A field in the form of a parallelogram has an area of $432 m^{2} .$ Find the length of its diagonal on which the perpendiculars drawn from two vertices on either side of it are $12 m$ long.

Area of the triangle $= \frac{1}{2} \times b \times h$
$\begin{matrix} = \frac{1}{2} \times x \times h_{1} + \frac{1}{2} \times x \times h_{2} = 432 = \frac{1}{2} \times x \times 12 + \frac{1}{2} \times x \times 12 = 432 12 x = 432 x = 36 \end{matrix}$
Hence, the length of diagonal is $36$ m.

A field in the form of a parallelogram has an area of 432m2. Find the length of its diagonal on which the perpendiculars drawn from two vertices on either side of it are 12m long.

Area of the triangle =12×b×h=12×x×h1+12×x×h2=432=12×x×12+12×x×12=43212x=432x=36Hence, the length of diagonal is 36 m.

A field in the form of a parallelogram has an area of $432 m^{2} .$ Find the length of its diagonal on which the perpendiculars drawn from two vertices on either side of it are $12 m$ long.

Area of the triangle $= \frac{1}{2} \times b \times h$
$\begin{matrix} = \frac{1}{2} \times x \times h_{1} + \frac{1}{2} \times x \times h_{2} = 432 = \frac{1}{2} \times x \times 12 + \frac{1}{2} \times x \times 12 = 432 12 x = 432 x = 36 \end{matrix}$
Hence, the length of diagonal is $36$ m.