SSC (English Medium) इयत्ता १० वी - Maharashtra State Board Question Bank Solutions for Geometry Mathematics 2

If ΔXYZ ~ ΔPQR then `"XY"/"PQ" = "YZ"/"QR"` = ?

Areas of two similar triangles are in the ratio 144: 49. Find the ratio of their corresponding sides.

In fig., TP = 10 cm, PS = 6 cm. `"A(ΔRTP)"/"A(ΔRPS)"` = ?

Ratio of corresponding sides of two similar triangles is 4:7, then find the ratio of their areas = ?

In fig. BD = 8, BC = 12, B-D-C, then `"A(ΔABC)"/"A(ΔABD)"` = ?

In fig., PM = 10 cm, A(ΔPQS) = 100 sq.cm, A(ΔQRS) = 110 sq.cm, then NR?

ΔPQS and ΔQRS having seg QS common base.

Areas of two triangles whose base is common are in proportion of their corresponding [______]

`("A"("PQS"))/("A"("QRS")) = (["______"])/"NR"`,

`100/110 = (["______"])/"NR"`,

NR = [ ______ ] cm

In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?

From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?

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)"`

Four alternative answers for the following question is given. Choose the correct alternative.

How many circles can drawn passing through three non-collinear points?

How many circles can be drawn passing through a point?

Choose the correct alternative:

______ number of tangents can be drawn to a circle from the point on the circle.

Choose the correct alternative:

In the figure ΔABC ~ ΔADE then the ratio of their corresponding sides is ______

Choose the correct alternative:

ΔPQR ~ ΔABC, `"PR"/"AC" = 5/7`, then

Choose the correct alternative:

∆ABC ∼ ∆AQR. `"AB"/"AQ" = 7/5`, then which of the following option is true?

Draw seg AB of length 9 cm and divide it in the ratio 3 : 2

∆ABC ~ ∆PBQ. In ∆ABC, AB = 3 cm, ∠B = 90°, BC = 4 cm. Ratio of the corresponding sides of two triangles is 7 : 4. Then construct ∆ABC and ∆PBQ

ΔRHP ~ ΔNED, In ΔNED, NE = 7 cm, ∠D = 30°, ∠N = 20° and `"HP"/"ED" = 4/5`. Then construct ΔRHP and ΔNED

ΔPQR ~ ΔABC. In ΔPQR, PQ = 3.6cm, QR = 4 cm, PR = 4.2 cm. Ratio of the corresponding sides of triangle is 3 : 4, then construct ΔPQR and ΔABC

Construct an equilateral ∆ABC with side 5 cm. ∆ABC ~ ∆LMN, ratio the corresponding sides of triangle is 6 : 7, then construct ΔLMN and ΔABC