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## AREA IN THE ANGLE TEST-1

TYT-AYT Geometry triangle area test-1 and solutions …

Question 1-) ABC is a right triangle, [AB] perpendicular [AC], [ED] parallel [BC], | AE | = 4 cm, | DC | = 3 cm; How many cm² is the area (EBD)?

Question 2-) ABC is a triangle, [DE] perpendicular [EC], | EC | = 6 cm, | DE | = 8 cm, | AD | = 5 cm, | BE | If = 10 cm; How many cm² is the area (ABED)?

Question 3-) In the figure, triangle ABC [AB] is divided into 4 equal parts, [BC] into 3 equal parts. Accordingly, what is the Area (DKL) / Area (ABC) ratio?

Question 4-) ABC is a triangle, point F [AE] and [DC] nine cut points, | DB | = 4 cm, | BE | = 6 cm, | EC | = 8 cm, if Area (DBEF) = Area (AFC); | AD | = How many cm is x?

Question 5-) ABC and FBD are a triangle, Area (AFE) = Area (ECD), | AF | = 6 cm, | BC | = 12 cm, | CD | If = 8 cm; | FB | = How many cm is x?

Question 6-) ABC is a triangle, m (ADB) = 60 °, | AE | = 7 cm, | BC | = 12 root if 3 cm; How many cm² is the area (ABEC)?

Question 7-) In the right triangle ABC, [DE] parallel [BC], | DE | = 6 cm, | AB | = 14 cm; How many cm² is the area (ADC)?

Question 😎 In the figure [AB] perpendicular [BC], [AD] perpendicular [AC], | AD | = | AC |, | AB | If = 12 cm; How many cm² is the area (ADB)?

### AREA IN ANGLE -TEST-1 SOLUTIONS

Solution: If we draw perpendicular from point D to edge BC, all angles of triangle ADE and HCD are equal. From the ratio of the similarity of the two triangles, the base x height value of the EDB triangle and the area of the EDB triangle are calculated.

Solution 2: The spacing right triangle is the 6 8 10 triangle. | DC | If we descend perpendicular to the base from corner A in ABC triangle, [DC] will be parallel, so similar triangles are formed. From the similarity ratio | AH | = 12 cm. Since the area of the two triangles is known, the shaded head area = 96-24 = 72 cm².

Solution 3: If we connect D with C point. Since the base of the triangle in DBC is divided into equal parts, the three triangles are equal. Area (DBC) = 3S. | BD | = 3 | AD | domain (ADC) = S. Area (DKL) / Area (ABC) = 1/4.

Solution: If we name the heights belonging to the side [BC] and [AB] with h1 and h2, write 2 equations from equal areas (S + S1), (S + S2) and divide by side, the value of x is found..

Solution 5: If we letter the equal areas, the area of triangle ABC is equal to the area of triangle FBD. Write the fields in terms of the two sides and the angle between them, if we equal each other we find the length.

Solution 6: The area of the convex quadrilateral ABEC in the figure is found by subtracting the small triangle (EBC) from the area of the greater triangle (ABC). To facilitate the solution.

You can find the proof of this formula here.

Solution: If we draw [DB] and [EB] in the right triangle ABC; In quadrilateral DB DBCE [DE], the area of two SPACING triangles parallel to [BC] (trapezoid) converges DEB triangle. The hatched area area becomes equal to the location of the AEBD convex quadrilateral. Area (AEBD) = (6.14) / 2 = 42 cm².

Solution: Let’s draw perpendicular to [AC] and [AD] from point B in the convex square in the figure. In AEBF rectangle | EB | = | AF | dir. In triangle ABC, 12² = y.x from the Euclidean relation, and Area (ADB) = 72 cm².[/vc_column_text][/vc_column][/vc_row]