Question

15. In a trapezium ABCD, seg AB || seg DC,
seg BDI seg AD, seg ACI seg BC, if AD = 15,
BC = 15 and AB = 25. Find A(O ABCD).
Note :1.By use pytha. theorem ​

Answer 1

Answer:

192 sq.units

Step-by-step explanation:

according according to Pythagoras theoremtriangleABD

AB²= AD² +BD²

BD²=AB²-AD²

BD=20

Now,

Area of triangle ABD = √S(S-a) (S-b) (S- c)

S= a+b+c/2S=30

Area of triangle =√S(S-a) (S-b) (S- c)

Area of triangle = 150 sq . units

DP = 300 /25 ......... (1/2*base *height)

DP=12

therefore height of trapezium is 12

Now according to pythagoras theoremIn triangle ADP

AD²=AP²+DP²

AP²=AD²-

AP²=224-144

AP=9

therefore AP= QB=9

CD=PQ=7

area of trapezium =1/2 * some of parallel sides * height =192 sq.units

hence area of trapezium ABCD is 192 sq.units

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Answer 2

Answer:

• In ∆ ADB, Using Pythagoras theorem :

⇒ AB² = AD² + BD²

⇒ (25)² = (15)² + BD²

⇒ (25)² - (15)² = BD²

• a² - b² = (a + b)(a - b)

⇒ (25 + 15)(25 - 15) = BD²

⇒ 40 × 10 = BD²

⇒ 400 = BD²

• Squaring root both sides

⇒ BD = 20

⠀⠀⠀⠀⠀───────────────

As ∆ ADB is a right angle triangle, and it is having two heights i.e. BD and DM.

• Area of right angle ∆ ADB :

⇒ Area with base AD = Area with base AB

⇒ ½ × Base × Height = ½ × Base × Height

⇒ ½ × AD × BD = ½ × AB × DM

• Cancelling ½ from both sides

⇒ AD × BD = AB × DM

⇒ 15 × 20 = 25 × DM

⇒ 300 = 25 × DM

⇒ 300/25 = DM

⇒ DM = 12

• In ∆ AMD, Using Pythagoras theorem :

⇒ AD² = DM² + AM²

⇒ (15)² = (12)² + AM²

⇒ (15)² - (12)² = AM²

⇒ (15 + 12)(15 - 12) = AM²

⇒ 27 × 3 = AM²

⇒ 81 = AM²

• Squaring root both sides

⇒ AM = 9

• MN = AB - (AM + BN) = 25 - (9 + 9) = 25 - 18 = 7 = CD

━━━━━━━━━━━━━━━━━━━━━

• MN = CD = 7
• AB = 25
• DM = 12

• Area of Trapezium ABCD :

⇢ Area = ½ × (Sum of parallel sides) × Height

⇢ Area = ½ × (CD + AB) × DM

⇢ Area = ½ × (7 + 25) × 12

⇢ Area = 32 × 6

⇢ Area = 192 sq units

∴ Hence, Area of trapezium is 192 sq units.