SSC (Marathi Semi-English) 10th Standard Board Exam [इयत्ता १० वी] - Maharashtra State Board Important Questions for Geometry Mathematics 2

In ΔABC, ∠ACB = 90°. seg CD ⊥ side AB and seg CE is angle bisector of ∠ACB.

Prove that: `(AD)/(BD) = (AE^2)/(BE^2)`.

The ratio of the areas of two triangles with the common base is 4 : 3. Height of the larger triangle is 2 cm, then find the corresponding height of the smaller triangle.

In ∆ABC, B – D – C and BD = 7, BC = 20, then find the following ratio.

`(A(triangleABD))/(A(triangleABC))`

In the given, seg BE ⊥ seg AB and seg BA ⊥ seg AD.

if BE = 6 and AD = 9 find `(A(Δ ABE))/(A(Δ BAD))`.

In the figure, ray YM is the bisector of ∠XYZ, where seg XY ≅ seg YZ, find the relation between XM and MZ. 

In the above figure, line l || line m and line n is a transversal. Using the given information find the value of x. 

Draw seg AB = 6.8 cm and draw perpendicular bisector of it. 

In the above figure, line AB || line CD || line EF, line l, and line m are its transversals. If  AC = 6, CE = 9. BD = 8, then complete the following activity to find DF. 

Activity :

`"AC"/"" = ""/"DF"`   (Property of three parallel lines and their transversal) 

∴ `6/9 = ""/"DF"`

∴ `"DF"  = "___"`

In the following figure, ray PT is the bisector of ∠QPR Find the value of x and perimeter of ∠QPR.

A roller of diameter 0.9 m and the length 1.8 m is used to press the ground. Find the area of the ground pressed by it in 500 revolutions.
`(pi=3.14)`

Draw the circumcircle of ΔPMT in which PM = 5.6 cm, ∠P = 60°, ∠M = 70°.

In ΔABC, B − D − C and BD = 7, BC = 20, then find the following ratio.

(i) `"A(ΔABD)"/"A(ΔADC)"`

(ii) `"A(ΔABD)"/"A(ΔABC)"`

(iii) `"A(ΔADC)"/"A(ΔABC)"`

Prove that, The areas of two triangles with the same height are in proportion to their corresponding bases. To prove this theorem start as follows:

1. Draw two triangles, give the names of all points, and show heights.
2. Write 'Given' and 'To prove' from the figure drawn.

From the information given in the figure, determine whether MP is the bisector of ∠KMN.

If ΔABC ∼ ΔDEF such that ∠A = 92° and ∠B = 40°, then ∠F = ?

If ΔABC ∼ ΔDEF, length of side AB is 9 cm and length of side DE is 12 cm, then find the ratio of their corresponding areas.

In the figure, PQ ⊥ BC, AD ⊥ BC. To find the ratio of A(ΔPQB) and A(ΔPBC), complete the following activity.

Given: PQ ⊥ BC, AD ⊥ BC

Now, A(ΔPQB)  = `1/2 xx square xx square`

A(ΔPBC)  = `1/2 xx square xx square`

Therefore, 

`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`

= `square/square`

In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR. 

Complete the proof by filling in the boxes.

solution:

In ∆PMQ,

Ray MX is the bisector of ∠PMQ.

∴ `("MP")/("MQ") = square/square` .............(I) [Theorem of angle bisector]

Similarly, in ∆PMR, Ray MY is the bisector of ∠PMR.

∴ `("MP")/("MR") = square/square` .............(II) [Theorem of angle bisector]

But `("MP")/("MQ") = ("MP")/("MR")`  .............(III) [As M is the midpoint of QR.] 

Hence MQ = MR

∴ `("PX")/square = square/("YR")`  .............[From (I), (II) and (III)]

∴ XY || QR   .............[Converse of basic proportionality theorem]

In ΔABC, ray BD bisects ∠ABC, A – D – C, seg DE || side BC, A – E – B, then for showing `("AB")/("BC") = ("AE")/("EB")`, complete the following activity:

Proof :

In ΔABC, ray BD bisects ∠B.

∴ `square/("BC") = ("AD")/("DC")`   ...(I) (`square`)

ΔABC, DE || BC

∴ `(square)/("EB") = ("AD")/("DC")`   ...(II) (`square`)

∴ `("AB")/square = square/("EB")`   ...[from (I) and (II)]

In the following figure, in ΔABC, ∠B = 90°, ∠C = 60°, ∠A = 30°, AC = 18 cm. Find BC.