Question

The sides of right triangles are calculated. Using this, calculate the perimeter of the
triangles given below.
∆BAD=
∆CAD=
AB=
BD=
CD=
BC=
AC=
Find the perimeter of triangle ABC​

Answer 1

Solution :-

In right angled ∆ADB, we have,

→ ∠ABD = 30°

→ AD = 1 cm

so,

→ sin 30° = AD/AB

→ (1/2) = 1/AB

→ AB = 2

and,

→ tan 30° = AD/BD

→ (1/√3) = 1/BD

→ BD = √3 cm

similarly, In right angled ∆ADC, we have,

→ ∠ACD = 30°

→ AD = 1 cm

so,

→ sin 30° = AD/AC

→ (1/2) = 1/AC

→ AC = 2 cm

and,

→ tan 30° = AD/CD

→ (1/√3) = 1/CD

→ CD = √3 cm

then, In ∆ABC we have,

→ AB = 2 cm

→ AC = 2 cm

→ BC = BD + DC = √3 + √3 = 2√3 cm

therefore,

→ Perimeter of ∆ABC = AB + AC + BC = 2 + 2 + 2√3 = 4 + 2√3 = 2(2 + √3) cm

Learn more :-

in triangle ABC seg DE parallel side BC. If 2 area of triangle ADE = area of quadrilateral DBCE find AB : AD show that B...

brainly.in/question/15942930

2) In ∆ABC seg MN || side AC, seg MN divides ∆ABC into two parts of equal area. Determine the value of AM / AB

brainly.in/question/37634605

Answer 2

Answer:

→ AM / AB = (2 - √2)/2 (Ans.)

Step-by-step explanation:

given ,

→ Area (∆BMN) = Area (Quad.MACN)

→ Area (∆BMN) / Area (Quad.MACN) = 1/1

so,

→ Area (∆BMN) / Area (∆BMN) + Area (Quad.MACN) = 1/(1 + 1) = 1/2

→ Area (∆BMN) / Area (∆BAC) = 1/2

now, in ∆BMN and ∆BAC, we have,

→ ∠BMN = ∠BAC (given that, DE || BC, so , corresponding angles .)

→ ∠BNM = ∠BCA (corresponding angles .)

then,

→ ∆BMN ~ ∆BAC (By AA similarity.)

now, we know that,

Ratio of areas of two similar ∆'s = Ratio of square of their corresponding sides.

therefore,

→ Area (∆BMN) / Area (∆BAC) = BM²/BA²

→ (1/2) = (BM/BA)²

square root both sides,

→ BM / BA = 1/√2

hence,

→ (BA - BM) / BA = (√2 - 1) / √2

→ AM / BA = (√2 - 1)/√2

→ AM / AB = ((√2 - 1)/√2) * (√2/√2)

→ AM / AB = √2(√2 - 1) / 2

→ AM / AB = (2 - √2)/2 (Ans.)