Question

The area of a rectangle is 216 cm square one side of the triangle measures 24 cm find the distance of the corresponding vertex from the side

Answer 1

ABCD is rectangle .
AC is diagonal
AB||CD, AB=CD
} ....... (Opposite side of rectangle is parallel & equal)
AD||BC ,AD=BC
In∆ABD & ∆BDC
angle ABD = angle BDC ..... (alternate angle)
angle ADB = angle CBD
AB = CD
∆ABD congruent ∆BDC. ....... (By ASA test)

BUT AB=CD = 24
Area of rectangle = length × breadth
216 = 24 × breadth
Breadth. = 216/24
Breadth. = 9
The distance between corresponding side is breadth is 9cm.

Answer 2

GIVEN :-

area of triangle = 216 cm²

base of triangle = 24 cm

TO FIND :-

height of triangle ( distance of the corresponding vertex from the side.)

SOLUTION :-

we know that the area of triangle

$\implies \boxed{ \rm{area \: of \: \triangle \: = \dfrac{1}{2} \times base \times hieght \: }}</p><p></p><p></p><p>$

so ,

$\implies \rm{ 216 \: = \: \dfrac{1}{2} \times 24 \times h}$

$\implies \rm{ 216 \: = \: 12 \times h}$

$\implies \rm{ h = \dfrac{216}{12} }$

$\implies \boxed { \boxed {\rm{ h = 18 cm }}}$

OTHER INFORMATION

Properties of triangles:

• If two triangles are similar, ratios of sides = ratio of heights = ratio of medians = ratio of angle bisectors = ratio of inradii = ratio of circum radii.

• Ratio of areas = b1h1/b2h2 = (s1)2/(s2)2 , where b1& h1 are the base & height of first triangle and b2& h2 are the base & height of second triangle. s1& s2 are the corresponding sides of first and second triangle respectively.

• The two triangles on each side of the perpendicular drawn from the vertex of the right angle to the largest side i.e. Hypotenuse are similar to each other & also similar to the larger triangle