Description : Factorise: a(square) + b(square) - 2ba + 2bc - 2ca -Maths 9th
Last Answer : Solution :-
Description : PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th
Last Answer : This is the sketch of the question but its hard to answer.
Description : in triangle abc if bd =1/3 bc then prove that 9(ad -Maths 9th
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Description : in triangle abc bd =1/3 bd then prove that 9(ad)^2=7(ab)^2 -Maths 9th
Description : If in equilateral triangle ABC, AD is perpendicular on BC then Prove that 3ABsquar=4ADsquare -Maths 9th
Description : ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th
Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)
Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th
Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........
Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th
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Description : Bisectors of angles A, B and C of a triangle ABC intersects its circumcircle at D, E and F respectively. Prove that angles of triangle DEF are 90° - A/2, 90° - B/2 and 90° - C/2. -Maths 9th
Last Answer : We have ∠BED = ∠BAD (Angles in the same segment) ⇒ ∠BED = 1/2∠A ...(i) Also, ∠BEF = ∠BCF (Angles in the same segment) ⇒ ∠BEF = 1/2∠C ...(ii) From (i) and (ii) ∠BED + ∠BEF = 1/2∠A + 1/2∠C ∠DEF ... ∠A + ∠C) ⇒ ∠DEF = 1/2(180° - ∠B) (Since, ∠A + ∠B + ∠C = 180°) ⇒ ∠DEF = 90° - 1/2∠B
Description : In triangle ABC, D and E are mid-points of the sides BC and AC respectively. Find the length of DE. Prove that DE = 1/2AB. -Maths 9th
Last Answer : First Find the points D and E by midpoint formula. (x₂+x₁/2 , y₂+y₁/2) For DE=1/2AB In ΔsCED and CAB ∠ECD=∠ACB and the ratio of the side containing the angle is same i.e, CD=1/2BC ⇒CD/BC=1/2 EC=1/2AC ⇒EC/AC=1/2 ∴,ΔCED~ΔCAB hence the ratio of their corresponding sides will be equal, DE=1/2AB
Description : If abc,=1 then -Maths 9th
Last Answer : Here is the answer I Have found. If you find it useful please upvote.
Description : In Fig. 4.6, if ABC and ABD are equilateral triangles then find the coordinates of C and D. -Maths 9th
Description : If a, b, c are the sides of a non-equilateral triangle, then the expression (b + c – a) (c + a – b) (a + b – c) – abc is -Maths 9th
Last Answer : answer:
Description : In an equilateral triangle ABC, the side BC is trisected at D. Then AD^2 is equal to -Maths 9th
Description : Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th
Description : Let ABC be a right angled triangle with AC as its hypotenuse. Then, -Maths 9th
Description : If a + b + c = 9 and ab + bc + ca = 23, then a3 + b3 + c3 – 3 abc = (a) 108 (b) 207 (c) 669 (d) 729 -Maths 9th
Last Answer : a+b+c=9 and a2+b2+c2=35 Using formula, (a+b+c)2=a2+b2+c2+2(ab+bc+ca) 92=35+2(ab+bc+ca) 2(ab+bc+ca)=81−35=46 (ab+bc+ca)=23 using formula, (a3+b3+c3)−3abc=(a2+b2+c2−ab−bc−ca)(a+b+c) a3+b3+c3−3abc=(35−23)×9=9×12=108
Description : If A (-2, 4), B (0, 0) and C (4, 2) are the vertices of triangle ABC, then find the length of the median through the vertex A. -Maths 9th
Last Answer : D=slid ht of BC D≅(20+4,20+2) =(2,1) ∴ Length of median = Light of AD =root(−2−2)2+(4−1)2=root42+32=5 hope it helps thank u
Description : The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th
Last Answer : area of parallelogram=2× area of triangle ABC =2×15=30sq cm theorem on area
Description : 2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal. -Maths 9th
Last Answer : Consider the following diagram- Here, it is given that AOB = COD i.e. they are equal angles. Now, we will have to prove that the line segments AB and CD are equal i.e. AB = CD. Proof: In triangles AOB ... ) So, by SAS congruency, ΔAOB ΔCOD. ∴ By the rule of CPCT, we have AB = CD. (Hence proved).
Description : If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th
Last Answer : Let AB and AC be two chords and AD be the diameter of the circle . Draw OL ⊥ AB and OM ⊥ AC . In ΔOLA and ΔOMA ∠ OLA = ∠ OMA = 90° OA = OA [common sides] ∠ OAL = ∠ OAM [given] . OLA ≅ OMA [by ASA congruency] OL =OM (by CPCT) Chords AM and AC are equidistant from O. ∴ AB = AC Hence proved.
Description : If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th
Last Answer : Let ABCD be quadrilateral with ab||cd Join be. In triangle abd and CBD, Angle abd=angle cdb(alternate angles) Anglecbd=angle adb(alternate angles) Bd=bd(common) Abd=~CBD by asa test Ad=BC by cpct Since ad ... c(from 1) Ad =bc(proved above) Triangle adc=~bcd by sas test Ac=bd by cpct Hence proved
Description : If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th
Last Answer : Prove that a3 +b3 +c3 – 3abc = -25
Description : If a pair of opposite sides of a cyclic quadrilateral are equal, then prove that its diagonals are also equal. -Maths 9th
Last Answer : Given Let ABCD be a cyclic quadrilateral and AD = BC. Join AC and BD. To prove AC = BD Proof In ΔAOD and ΔBOC, ∠OAD = ∠OBC and ∠ODA = ∠OCB [since, same segments subtends equal angle to the circle] AB = BC [ ... is DOC on both sides, we get ΔAOD+ ΔDOC ≅ ΔBOC + ΔDOC ⇒ ΔADC ≅ ΔBCD AC = BD [by CPCT]
Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th
Last Answer : According to question prove that P, Q, R and D are concyclic.
Description : If two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle. -Maths 9th
Last Answer : Given In a circle AYDZBWCX, two chords AB and CD intersect at right angles. To prove arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle. Construction Draw a diameter EF parallel to CD having centre M. Proof ... (i) arc ECXA = arc EWB [symmetrical about diameter of a circle] arc AF = arc BF (ii)
Description : if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th
Last Answer : Since complete line is AC and B is point on it. therefore, AC is divide into 2 parts AB&BC. therefore, AC=AB+BC
Last Answer : AB=AC-BC BC =AC-AB AB+BC=AB HENCE PROVED
Description : If P, Q and R are three points on a line and Q is between P and R,then prove that PR - QR= PQ. -Maths 9th
Description : In Fig.5.6, if AC = BD, then prove that AB = CD. -Maths 9th
Last Answer : if equals are subtracted to equals ,the remainders are equal subtract bc on both sides ab-bc=bd-bc ab=cd hence proved
Description : If a point O lies between two points P and R such that PO=OR then prove that PO= 1/2PR. -Maths 9th
Last Answer : THINGS WHICH ARE COINCIDE WITH EACH OTHER ARE EQUAL TO ONE ANOTHER PO+OR=PR 2PO=PR PO=OR PO=1/2PR HENCE PROVED
Description : In Fig. 6.14, if x+y = w+z,then then prove that AOB is a line. -Maths 9th
Description : Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then two triangles are congruent. -Maths 9th
Description : If AOB is a diameter of a circle [Fig. 10.8] and C is a point on the circle, then prove that AC* +BC*=AB*. -Maths 9th
Last Answer : Solution :- As, ∠ C = 90° (Angle in the semicircle) ∴ AC2 + BC2 = AB2 (By Pythagoras Theorem)
Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th
Last Answer : Solution :- Let AB be a chord of a circle having centre at O. Let PQ be the perpendicular bisector of the chord AB intersect it say at M. Perpendicular bisector of the chord passes through the centre of the circle,i. ... = PM (Common) ∴ △APM ≅ △BPM (SAS) PA = PB (CPCT) Hence, arc PXA ≅ arc PYB
Description : Prove that if the opposire anles of a parallelogram are equal then it is a rectangle -Maths 9th
Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th
Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB
Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th
Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.