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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]