Question

For the parallelogram ABCD the extensions of the angle bisectors AG and BH intersect at point P. Find the area of the parallelogram, if DG = GH, AB = 39 in, GP = 12 in.

Answer 1

The area of a parallelogram is 360 in.²

Explanation:

Where DG = GH

GP = 12 in.

AB = 39 in.

∠DAB + ∠ABC = 180° (Adjacent angles of a parallelogram)

Whereby ∠DAB is bisected by AG and ∠ABC is bisected by BH

Therefore, ∠GAB + ∠HBA = 90°

Hence, ∠BPA = 90° (Sum of interior angles of a triangle)

`cos(\angle GAB) = (AP)/(AB) = (AP)/(39) = (GP)/(GH) =(12)/(GH)`

We note that ∠AGD = ∠GAB (Alternate angles of parallel lines)

∴ ∠AGD = ∠AGD since ∠AGD = ∠GAB (Bisected angle)

Hence AD = DG (Side length of isosceles triangle)

The bisector of ∠ADG is parallel to BH and will bisect AG at point Q

Hence ΔDAQ ≅ ΔDGQ ≅ ΔGPH and AQ = QG = GP

Hence, AP = 3 × GP = 3 × 12 = 36

`cos(\angle GAB) = (AP)/(AB) = (36)/(39)`

`\angle GAB = cos^(-1) \left ((36)/(39) \right )`

∠GAB = 22.62°

`cos(\angle GAB) = (36)/(39) = (12)/(GH)`

`GH = (39)/(36) * {12}`

GH = 13 in.

∴ AD 13 in.

BP = 39 × sin(22.62°) = 15 in.

GH = √(GP² + HP²)

∠DAB = 2 × 22.62° = 45.24°

The height of the parallelogram = AD × sin(∠DAB) = 13 × sin(45.24°)

The height of the parallelogram = 120/13 = 9.23 in.

The area of a parallelogram = Base × Height = (120/13) × 39 = 360 in.²

Answer 2

360 in2

Explanation:

The figure ABCD is a parallelogram, so the side AB is parallel to the side DC, and that means the angle mBAG is equal the angle mHGP.

The angles mHGP and mAGD are vertically opposite angles, so mHGP = mAGD.

The angle mDAG is equal the angle mBAG, so mDAG = mBAG = mHGP = mAGD

The angles mDAG and mAGD are equal, so the triangle DAG is isosceles, that means the side DG is equal the side DA.

In the same way, we can find that the angle mCBH is equal the angle mBHC, so triangle BCH is isosceles, meaning that side BC (which is equal side AD) is equal side CH, so we have that:

DG + GH + HC = AB = 39

GH + GH + GH = 39

GH = 13

in the parallelogram we have the property:

mDAB + mABC = 180°

We have that mGAB = mDAB/2 and mABH = mABC/2, so using this information for the sum of internal angles of triangle APB, we have:

mGAB + mABH + mBPA = 180

mDAB/2 + mABC/2 + mBPA = 180

(180/2) + mBPA = 180

mBPA = 90°

If the triangle APB is a right triangle, we can find the cosine of the angle mHGP:

cos(mHGP) = GP / GH = 12 / 13

Then, we can find the sine of mHGP using:

sin(mHGP)^2 + cos(mHGP)^2 = 1

sin(mHGP)^2 + 144/169 = 1

sin(mHGP)^2 = 25/169

sin(mHGP) = 5/13

If we draw the height of the triangle GPH relative to the side GH, we have:

sin(mHGP) = height / GP

5 / 13 = height / 12

height = 60/13

The triangles GPH and APB are similar (using the case angle-angle), so we can use this relation to find the height 'h2' of the triangle APB:

height / (h2+height) = GH / AB

(60/13) / (h2 + (60/13)) = 13/39 = 1/3

180 / 13 = h2 + (60/13)

h2 = 120/13

Then the area of the parallelogram is:

Area = AB * h2 = 39 * 120/13 = 360 in2