goi a,b,c la cac canh cua 1 tam giac co 3 duong cao tuong ung la ha,hb,hc. cmr (a+b+c)^2/ha^2+hb^2+hc^2 lon hon hoac bang 4
mot tam giac co do dai 3 canh la a,b,c.goi ha,hb,hc theo thu tu la do dai 3 duong cao tuong ung.chung minh rang,neu ha=4,hb=5,hc=8 thi a2>.b2+c2
CHO DO DAI CAC CANH TAM GIAC ABC LA ;B;CVA DUONG CAO TUONG UNG LA HA;HB;HC BIET CHU VI TAM GIAC LA 94CM VA CAC DUONG CAO THEO THU TU TREN TI LE VOI 3;4;5 TIM A;B;C
goi ha,hb,hc la 3 duong cao cua mot tam giac.c/m (1/ha)<(1/hb+1/hc)
cho tam giac abc vuong tai a , duong cao ah . goi d,e lan luot la chan cac duong vuong goc ha tu h xuong ab va ac . goi m,n la trung diem cua hc va hb .
a, c/m tu giac HDAE LA HINH CHU nhat
b,tinh MD bt ab=15cm ; ac=20cm
c,tu giac DEMN la hinh gi ? vi sao?
Cho tam giac ABC CMR neu ha=3,hb=4,hc=5 thi a mu 2 lon hon b mu 2 cong c mu 2
cho tam giac ABC vuong tai A, AB< AC , AH la duong cao
a)chung minh Tam giac HAC va Tam giac ABC đồng dạng
b) cm HA2= HB . HC
c goi D, E lan luot la trung diem cua AB, BC chung minh CH . CB = 4DE2
cho tam giac ABC vuong tai A,duong cao AH 1,HB=2,HC=6 a,tinh AB, AH b,goc C va gocB cua tam giac ABC 2,goi E,Flan luot la hinh chieu cua H tren AB,AC.chung minhAB mu 3 tren AC mu 3=BE tren CF. giai cau 2 ho minh nhe canh nhanh cang tot thank ae.
1:
a: \(AH=\sqrt{2\cdot6}=2\sqrt{3}\left(cm\right)\)
\(AB=\sqrt{2\cdot8}=4\left(cm\right)\)
b: Xét ΔABC vuông tại A có sin C=AB/BC=1/2
nên góc C=30 độ
=>góc B=60 độ
2: \(\dfrac{BE}{CF}=\dfrac{BH^2}{AB}:\dfrac{CH^2}{AC}\)
\(=\dfrac{BH^2}{AB}\cdot\dfrac{AC}{CH^2}\)
\(=\left(\dfrac{BH}{CH}\right)^2\cdot\dfrac{AC}{AB}=\dfrac{AB^4}{AC^4}\cdot\dfrac{AC}{AB}=\dfrac{AB^3}{AC^3}\)
Goi a, b, c la do dai ba canh cua tam giac nhon \(h_a,h_b,h_c\)lan luot la ba duong cao tuong ung.
cmr\(\frac{h^2_a+h^2_b+h_c^2}{\left(a+b+c\right)^2}\)<1/4
Cho tam giac ABC nhon co ba duong cao ha hb hc cmr 1/ha =1/ hb + 1/hc thi tam giac ABC vuong tai A