In triangle PQR length of the side QR is less than twice the length of the side PQ by 2 cm. Length of the side PR exceeds the length of the side PQ by 10 cm. The perimeter is 40 cm. The length of the smallest side of the triangle PQR is :
In the given figureΔABO ∼ ΔDCO. IfCD = 2cm, AB = 3 cm, OC = 3.2 cm, OD = 2.4 cm, then
In the given figure, AB || DE and BD || EF. Then
Two isosceles triangles have equal angles and their areas are in the ratio 16 : 25. Then, the ratio of their corresponding heights is
Let two isosceles triangles be ΔABC and ΔDEF And their corresponding altitudes are AM and DN.
Since the corresponding angles of both triangles are equal.
If PQR is an isosceles triangle and M is a point on QR such that PM⊥QR,then
Since, in a triangle the sum of squares of any wo sides is equal to twice the square of half of the third side together with twice the square of the median bisecting it.
If in two triangles ABC and DEF,
If in two triangles ABC and DEF,
ΔFDE∼ΔCAB because for similarity, all sides should be in proportion.
In trapezium ABCD, if AB||DC, AB = 9 cm, DC = 6 cm and BD = 12 cm, then BO is equal to
In an equilateral triangle ABC, if AD⊥BC,then
Since the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
If ΔABC ∼ ΔPQR such that AB = 1.2 cm, PQ = 1.4 cm, then
Two poles of height 8 m and 13 m are standing 12 m apart. The distance between their tops is
AB = 13 m Therefore, the distance between their tops is 13 m.
If ΔABC ∼ ΔEDF and ΔABC is not similar to ΔDEF, then which of the following is not true?
If ΔABC ∼ ΔEDF and ΔABC is not similar to ΔDEF,ΔDEF,then BC. DE = AB .EF
Here according to given condition, BC. DE ≠ AB .EF
In the adjoining figure AC || BD. If AC = 6 cm, AE = 3 cm, EB = 4 cm, ED = 8 cm, then CE and BD are respectively
Given: AC || BD.and AC = 6 cm, AE = 3 cm, EB = 4 cm, ED = 8 cm,
In triangles ACE and DEB, ∠AEC = ∠DEB [Vertically opposite angles] ∠ECA = ∠EDB [Alternate angles as
If ΔABC ∼ ΔDEF and EF = 1/3BC, then ar(ΔABC):(ΔDEF) is
A semicircle is drawn on AC. Two chords AB and BC of length 8 cm and 6 cm respectively are drawn in the semicircle. What is the measure of the diameter of the circle?
The diameter of circle is AC. Here ∠ABC is angle of semicircle.
AB ∴ ∠ ABC = 90° ∴ Δ ABC is aright angled triangle.
The length of the second diagonal of a rhombus whose side is 5 cm and one of the diagonal is 8 cm is
Let ABCD be a rhombus.
One diagonal BC = 8 cm and dise BC 5 cm
Since diagonals of a rhombus bisect each other at 90∘
In the given figures the measures of ∠P and ∠R are respectively
Since, Sides of ΔABC and ΔPQR are proportional, therefore, .
∴ ∠A = ∠Q, ∠B = ∠R and ∠C = ∠P
= 20° and ∠R
In the given figure, AP is equal to
In triangles APB and CPD,
∠APB = ∠CPD [Vertically opposite angles] ∠BAP = ∠ACD [Alternate angles as AB||CD]
In the given figure if
In ΔPOQ and ΔCOB, ∠PQO = ∠QBO
[Alternate angles] ∠POO = ∠COB [Vertically Opposite]
ABC is a right triangle right angled at B. BD is the altitude through B. If BD = 4 cm and AD = 3 cm then AC is equal to
The length of the side of a rhombus whose diagonals are of lengths 24 cm and 10 cm is
Let ABCD be the required rhombus whose diagonals are AC = 10 cm and BD = 24 cm. Let AC and BD intersect each other at O.
Then AB = BC = CD = DA
Also AO = CO = 10/2 = 5 cm
And BO = DO = 24/2 = 12 cm
In the given figure if ∠ADE = ∠ABC,then CE is equal to
In the given figure if AT = AQ = 6, AS = 3, TS = 4, then
The areas of two similar triangles are 100cm2 and 49 cm2. If the altitude of the larger triangle is 5 cm, then the corresponding altitude of the other triangle is equal to
Let the two similar triangles be ΔABC and ΔDEF such that ar(ΔABC) = 100 cm2 and ar(ΔDEF) = 49 cm2.
Let AM and DN be the respective altitudes of ΔABC and ΔDEF.
Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding altitudes
A man goes 15 m due west and then 8 m due north. How far is he from the starting point ?
Using Pythagoras Theorem,
AC2 = AB2 + BC2 = 64 + 225 = 289 8 m ⇒ AC 17 cm
If D is a point on side BC of ΔABC such that BD = CD = AD, then