Question

Explore Activity 2: Yes you Can Do IT!​

Answer 1

$\bold{\huge{\underline{ Solution }}}$

Given :-

• Here, we have given one isosceles trapezoid that is MATH
• In the given trapezoid, MA and HT are the bases and LV is a median
• The length of MA = 3y - 2 , HT = 2y + 4 and LV = 8.5

Part 1 :-

• The value of y = 3

Part 2 :-

Here, we have to

• Find the value of y

We know that the,

Length of the median of trapezium

$\sf{ m = }{\sf{\dfrac{1}{2}}}{\sf{ (a + b) }}$

• Here, a and b are the parallel sides of the trapezium.

Subsitute the required values in the above formula :-

$\sf{ LV = }{\sf{\dfrac{1}{2}}}{\sf{ [MA + HT]}}$

$\sf{ 8.5 = }{\sf{\dfrac{1}{2}}}{\sf{[ (3y - 2) + (2y + 4)]}}$

$\sf{ 8.5 = }{\sf{\dfrac{1}{2}}}{\sf{ [3y - 2 + 2y + 4]}}$

$\sf{ 8.5 = }{\sf{\dfrac{1}{2}}}{\sf{ [5y + 2] }}$

$\sf{ 8.5 {\times}2 = 5y + 3 }$

$\sf{ 17 = 5y + 2 }$

$\sf{ 5y = 17 - 2 }$

$\sf{ 5y = 15 }$

$\bold{ y = 3 }$

Hence, The value of y is 3

Part 3 :-

Here, we have to find the length of MA and HT

• We have, MA = 3y - 2
• HT = 2y + 4

For MA

$\sf{ MA = 3y - 2}$

Subsitute the value of y

$\sf{ MA = 3(3) - 2}$

$\sf{ MA = 9 - 2 }$

$\bold{ MA = 7 }$

For HT

$\sf{ HT = 2y + 4}$

Subsitute the value of y

$\sf{ MA = 2(3) + 4 }$

$\sf{ MA = 6 + 4 }$

$\bold{ MA = 10 }$

Hence, The length of MA and HT are 7 and 10 .