Respuesta :
Point R is the midpoint of FE¯¯¯¯¯ , so the coordinates of point R are (3a, b).
In △DEF , the length of the base, DF¯¯¯¯¯ , is
4a, and the height is 2b, so its area is
1/2×4a×2b = 4ab.
In △QRP , the length of the base, QR¯¯¯¯¯ , is
3a-a = 2a, and the height is b, so its area is 1/2×2a×b = ab .
Comparing the expressions for the areas proves that the area of the triangle created by joining the midpoints of an isosceles triangle is one-fourth the area of the larger isosceles triangle.
In △DEF , the length of the base, DF¯¯¯¯¯ , is
4a, and the height is 2b, so its area is
1/2×4a×2b = 4ab.
In △QRP , the length of the base, QR¯¯¯¯¯ , is
3a-a = 2a, and the height is b, so its area is 1/2×2a×b = ab .
Comparing the expressions for the areas proves that the area of the triangle created by joining the midpoints of an isosceles triangle is one-fourth the area of the larger isosceles triangle.
Answer: We can fill the boxes with help of below explanation.
Explanation: According to the given figure,
Triangle DEF is an isosceles triangle with sides DE, EF and FD.
And, where DE=EF and P,R and Q are the midpoints of the sides DF, EF and DE respectively.
Theorem: Area of a triangle created by joining the midpoints of an isosceles triangles is one-fourth the area of the isosceles triangle.
Proof:
Point Q is the midpoint of DE, so the coordinates of point Q are (a, b). (because mid point of a line segment always has coordinates half of the sum of corresponding coordinates of end points. )
Point R is the midpoint of FE , so the coordinates of point R are (
3a,b).
In triangle DEF , the length of the base, DF is 4a and the height is 2b, So, its area is 4ab.( because, Area of a triangle= 1/2×base×height
In triangle QRP , the length of the base, QR is 2a and the height is b,
So, its area is ab.
On comparing the above two expression, we found that, area of triangle DEF =4( area of triangle QRP)