## QUADRILATERALS TEST-1

TYT-AYT Geometry issues quadrilaterals test-1 and solutions …

Question 1) Quadrilateral ABCD, | AD | = | DC | = | CB |, if the measure of angle A is 125 °, and the measure of angle D is 60 °; The measure of angle C = how many degrees is x?

Question 2) If the quadrilateral ABCD is the diagonal of [AC] and [DB], m (BAC) = 60 °, m (CAD) = 50 °, m (ADB) = 35 °, m (BDC) = 30 °; How many degrees is m (CBD) = x?

Question 3) ABCD quadrilateral, [BD] bisector, [DC]; Perpendicular to [BC], if | AB | = 9 cm, | DC | = 12 cm, | BC | = 14 cm; | AD | = how many cm is x?

Question 4) In quadrilateral ABCD; The midpoints of the edges where K, L, M, N are located, If Area (DKN) = 6 cm², Area (ALK) = 10 cm², Area (LBM) = 16 cm²; Area (KLMN) + Area (NMC) = how many cm² is x + y?

## QUADRILATERALS** TEST-1 SOLUTIONS**

Solution: Let’s connect point A with point C.m (ADC) = 60 ° and | AD | = | DC | Since ACD triangle is equilateral, m (BAC) = 125 ° -60 ° = 65 °. In ABC triangle | CB | = | AC | m (BAC) = m (CBA) = 65 °. Since m (ACB) = 50 ° from the sum of the interior angles of the triangle, m (DCB) = 110 ° is found

Solution: [AE] on the [BC] edge; If we take a point E perpendicular to [BD] and join it to point D, ABED becomes deltoid. | AD | = | DE | = x and | EC | = 5 cm. DEC right triangle will be 5-12-13 triangle | AD | = x = 13.

Solution: The midpoints of the sides K, L, M, N are the parallelograms KLMN. Let’s draw the diagonal of the quadrilateral ABCD [AC]. Since the similarity ratio of similar triangles is 1/4, Area (ACD) = 6.4 = 24 cm², A (ABC) = 16.4 = It becomes 64 cm². From here, the area of the ABCD quadrilateral is 88 cm². 88 = 6 + 10 + 16 + x + y, x + y = 56 cm².